Physics · Gravitation · NEET
r is the full distance from the CENTRE of the Earth, not the height above the ground. If a satellite is at height h above the surface, then r = R + h, where R is Earth's radius (about 6400 km). A very common NEET mistake is to put only h into T = 2π√(r³/GM). Always add the radius first. For a satellite skimming the surface, h ≈ 0, so r ≈ R.
From T = 2π√(r³/GM), a larger r gives a larger T. Physically, a higher orbit is bigger (more distance to cover) and the orbital speed is also smaller there (v = √(GM/r) drops as r grows). Slower speed plus a longer path both make the trip take more time. This is Kepler's third law in action: T² is proportional to r³.
They are connected by simple circular motion: the satellite travels one circumference 2πr in one period T at speed v. So T = 2πr / v. Since v = √(GM/r), substituting gives T = 2πr / √(GM/r) = 2π√(r³/GM). If a question gives you v, you can jump straight to T = 2πr/v without re-deriving everything.
For a near-surface satellite, r ≈ R, so T = 2π√(R³/GM). Putting R = 6.4×10⁶ m and GM = gR² gives T = 2π√(R/g) ≈ 84 minutes (about 1.4 hours). Writing M in terms of mean density d gives the neat NEET result T² = 3π/(Gd), which depends only on Earth's density, not its size.
Use the ratio form of Kepler's third law: T₂/T₁ = (r₂/r₁)^(3/2). All the constants (G, M, 4π²) cancel, so you never need their values. Just plug in the two centre-distances r₁ = R + h₁ and r₂ = R + h₂ as multiples of R. This is exactly how the NEET 2019 '24 h at height 6R' problem is solved.
A satellite has period 24 h at height 6R from the earth's surface. The period of another satellite at height 2.5R from the surface is:
A satellite orbits just above the earth's surface with period T. If d is the mean density of the earth and G the gravitational constant, the quantity 3π/(Gd) represents:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
T = 2π√(r³ / GM), where r = R + h is the distance from Earth's centre, G is the gravitational constant, and M is Earth's mass. It can also be written T = 2πr / v using the orbital velocity v = √(GM/r).
No. The satellite's own mass cancels out, exactly like in free fall. T depends only on the orbit radius r and the central body's mass M. A heavy and a light satellite in the same orbit have the same period.
A geostationary satellite has a period of 24 hours (one sidereal day, 23 h 56 min) so it stays fixed above one point on the equator. Setting T = 24 h in the formula gives an orbit radius of about 42,000 km from Earth's centre, roughly 36,000 km above the surface.
Square the formula: T² = 4π²r³/(GM). Since 4π²/(GM) is a constant for one planet, T² is directly proportional to r³. So the satellite period formula is just Kepler's third law with the constant filled in.
The minimum period is for an orbit skimming the surface (r ≈ R), which gives T = 2π√(R/g) ≈ 84 minutes. No satellite can orbit Earth faster than this without hitting the surface.
